Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
04 Mar 2012, 16:19
Kudos
Bookmarks
While reviewing advanced quant and doing some questions on the forum, I realiized that my understanding of inequalities isnt as good as I thought. Here are few things I ran into that I am curious about...
1) Is |a| + |b| > |a + b| ?
Usually I break abs value questions into the possible cases when i see something like this: 4 > |a + b|. I rewrite with 4 as a pos and neg solution. With |b + c| > |a + b| i do the all positive case and the switch the signes on one side case. However, with |a| + |b| > |a + b| or even this |a| + |b| > 4, Im not sure how to break it down...
2) Not sure how to spell this out in a forum post, but in advanced gmat quant they use a numer line approach where you take the quadratic values that set the equation equal to zeros and use them to divide the number line. You then put a +/- for the different segments. This seems cleaner than listing and testing number cases. Is there a disadvantage to is I am not seeing?
3) Two parter. A) is there a way to factor a + a^-1? This comes from DS question #4 in chater #4 of advanced quant (given below) and B) what else did I do wrong on the question
? (the one below).
Is a + a^-1 > 2?
1) a>0
2) a<1
I manipulated to (a+1)/a > 2 then took pos and neg cases since we dont know the sign of a from the base a>1 or a<1.
Statement 1 s then not enough, and Statement 2 matches one of the cases i found from the base so I picked B. OA is C. Clearly I did something wrong...
1) Is |a| + |b| > |a + b| ?
Usually I break abs value questions into the possible cases when i see something like this: 4 > |a + b|. I rewrite with 4 as a pos and neg solution. With |b + c| > |a + b| i do the all positive case and the switch the signes on one side case. However, with |a| + |b| > |a + b| or even this |a| + |b| > 4, Im not sure how to break it down...
2) Not sure how to spell this out in a forum post, but in advanced gmat quant they use a numer line approach where you take the quadratic values that set the equation equal to zeros and use them to divide the number line. You then put a +/- for the different segments. This seems cleaner than listing and testing number cases. Is there a disadvantage to is I am not seeing?
3) Two parter. A) is there a way to factor a + a^-1? This comes from DS question #4 in chater #4 of advanced quant (given below) and B) what else did I do wrong on the question
? (the one below).Is a + a^-1 > 2?
1) a>0
2) a<1
I manipulated to (a+1)/a > 2 then took pos and neg cases since we dont know the sign of a from the base a>1 or a<1.
Statement 1 s then not enough, and Statement 2 matches one of the cases i found from the base so I picked B. OA is C. Clearly I did something wrong...
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
09 Mar 2012, 18:35
Kudos
Bookmarks
Hi, there. I'm happy to help a bit with this.
QUESTION #1
Yes, it's true that, as long as a=/0 and b=/0, then |a + b| =< |a| + |b|. That's an actual math rule. First of all, notice it's not a "less than" sign, but a "less than or equal to" sign --- that's very important. This idea is closely related to something Geometry known as the Triangle Inequality. The basic logic is ---if a and b are both positive, or both negative, then the two sides will be the same, but if a and b have opposite sign, they will get subtracted on the left side, while their absolute values will be added on the right side. Example |(-3) + 5| = |2| = 2, but |(-3)| + |5| = 3 + 5 = 8. Does that make sense?
Caveat: this is always true, but not necessarily all that useful in problem-solving. If you have a DS question asking explicitly about comparing |a + b| to |a| + |b|, then this formula would be relevant, but in a general absolute value question, you are much better off enumerating cases, as you say you do anyway.
QUESTION #2
Yes, using a number line to mark +/- cases is the standard and most widely used method of approaching quadratic inequalities. It is a 100% valid and legitimate approach. If this method makes sense to you, there is no disadvantage to using it every single time.
QUESTION #3
OK, a DS question.
Prompt: Is a + a^-1 > 2?
Here's an algebraic approach. The big idea is --- when you have an x + 1/x, that is, an x to the +1 and an x to the -1, those are exponents that differ by two --- that means, in hidden form, you are dealing with a quadratic. That's what I will bring out in the solution below.
Subtract 2 from both sides:
a - 2 + 1/a > 0
Factor out an 1/a
(1/a)*(a^2 - 2a + 1) > 0 we have the square of a binomial!
(1/a)*(a - 1)^2 > 0
So, of these two factors, (1/a) is positive when a > 0 and negative when a < 0. By contrast, (a - 1)^2 is positive for all real number a EXCEPT when a = 1 --- that makes the expression equal to zero.
Statement #1: a > 0
The expression will be positive for all values of a except at a = 0, in which case the left side equals zero, and the inequality is false. So, a continuous infinity of possible values of a would make the prompt question true, and only one would make it false, but that's enough to say we can't answer the prompt definitively. Statement #1 by itself is insufficient.
Statement #2: a <1
The good news is --- the dreaded value of a = 1 is avoided, so the (a - 1)^2 is always positive. BUT, now a can be less than zero, and if a is negative, then 1/a is also negative, and that would make the inequality false. Again, we don't have a definitive answer. Statement #2 by itself is insufficient.
Statements #1 & #2 combined:
Now, we know 0 < a < 1. We avoid the dreaded a = 1 value, so (a - 1)^2 is always positive, and because a has to be greater than zero, we know that 1/a is also always positive. Therefore, the product of those two terms is always positive, and the inequality is always true. Combined, the statements are sufficient.
I realize this is probably a very different approach from the one you took to solve this DS. If you'd like to post your work, I can help you figure out where you went wrong and what mistakes to avoid in the future.
Here's another related GMAT DS question, for more practice.
https://gmat.magoosh.com/questions/999
The question at that link, when you submit your answer, will be followed by a page with a complete video explanation.
Does everything I've said this make sense? Please let me know if you have any questions on what I've said, or if I can help you in any other way.
Mike
QUESTION #1
Yes, it's true that, as long as a=/0 and b=/0, then |a + b| =< |a| + |b|. That's an actual math rule. First of all, notice it's not a "less than" sign, but a "less than or equal to" sign --- that's very important. This idea is closely related to something Geometry known as the Triangle Inequality. The basic logic is ---if a and b are both positive, or both negative, then the two sides will be the same, but if a and b have opposite sign, they will get subtracted on the left side, while their absolute values will be added on the right side. Example |(-3) + 5| = |2| = 2, but |(-3)| + |5| = 3 + 5 = 8. Does that make sense?
Caveat: this is always true, but not necessarily all that useful in problem-solving. If you have a DS question asking explicitly about comparing |a + b| to |a| + |b|, then this formula would be relevant, but in a general absolute value question, you are much better off enumerating cases, as you say you do anyway.
QUESTION #2
Yes, using a number line to mark +/- cases is the standard and most widely used method of approaching quadratic inequalities. It is a 100% valid and legitimate approach. If this method makes sense to you, there is no disadvantage to using it every single time.
QUESTION #3
OK, a DS question.
Prompt: Is a + a^-1 > 2?
Here's an algebraic approach. The big idea is --- when you have an x + 1/x, that is, an x to the +1 and an x to the -1, those are exponents that differ by two --- that means, in hidden form, you are dealing with a quadratic. That's what I will bring out in the solution below.
Subtract 2 from both sides:
a - 2 + 1/a > 0
Factor out an 1/a
(1/a)*(a^2 - 2a + 1) > 0 we have the square of a binomial!
(1/a)*(a - 1)^2 > 0
So, of these two factors, (1/a) is positive when a > 0 and negative when a < 0. By contrast, (a - 1)^2 is positive for all real number a EXCEPT when a = 1 --- that makes the expression equal to zero.
Statement #1: a > 0
The expression will be positive for all values of a except at a = 0, in which case the left side equals zero, and the inequality is false. So, a continuous infinity of possible values of a would make the prompt question true, and only one would make it false, but that's enough to say we can't answer the prompt definitively. Statement #1 by itself is insufficient.
Statement #2: a <1
The good news is --- the dreaded value of a = 1 is avoided, so the (a - 1)^2 is always positive. BUT, now a can be less than zero, and if a is negative, then 1/a is also negative, and that would make the inequality false. Again, we don't have a definitive answer. Statement #2 by itself is insufficient.
Statements #1 & #2 combined:
Now, we know 0 < a < 1. We avoid the dreaded a = 1 value, so (a - 1)^2 is always positive, and because a has to be greater than zero, we know that 1/a is also always positive. Therefore, the product of those two terms is always positive, and the inequality is always true. Combined, the statements are sufficient.
Answer = C
I realize this is probably a very different approach from the one you took to solve this DS. If you'd like to post your work, I can help you figure out where you went wrong and what mistakes to avoid in the future.
Here's another related GMAT DS question, for more practice.
https://gmat.magoosh.com/questions/999
The question at that link, when you submit your answer, will be followed by a page with a complete video explanation.
Does everything I've said this make sense? Please let me know if you have any questions on what I've said, or if I can help you in any other way.
Mike
11 Mar 2012, 17:52
Kudos
Bookmarks
mikemcgarry
Thanks for the thoughful reply, Mike.
1) Thanks for the insight here. Is the cleanest way to test an abs value problem with multiple variables to just plug in numbers and check? It seems tough algebraically...
2) I'll keep using the line then. Thanks,
3) Your feedback on recognizing this as a quadratic helped a ton. I think the problem I keep running into with this type of problem is the rule set for switching the inequality. I think I am confusing inequalities with abs values and those without. Now that I am factoring properly, here is how I solved it. Instead of factoring out the 1/a as you did, I multiplied the a through and made equations for the positive and negative cases (here is where confusion sometimes sets in) and got a^2-2a>1 and a^2-2a<1. From there I just solved as you did using the number line concept.
When putting together the inequalities for positive and negative cases, if there are no abs values, you multiple through the same as with the positive case, but just switch the sign, right? And with abs values, you solve once for the postive case and for the negagtive you slap a minus on everything that was in the brackets, flippling the sign only if you multiple the other side by a negative, correct? It seems like my brain fluctuates begin "getting it" ad "not". This type of problem seems to be my only weakness in quant...
Thanks again for the help! Kudos!
